ForceWhat kinds of forces could be expected when ignition of a fully filled combustion chamber occurs? How much energy is generated in such a optimal ignition? What would the peak pressure be just after ignition? So, what dangers could this pose?To be able to calculate these forces, the engine displacement has to be known. The displacement is the measure of how much air the engine moves in one stroke. It is important to note that this volume is not equal to the chamber volume. The chamber volume is the total volume of the combustion chamber when the piston is at its lowest possible position (BDC or bottom dead center). The displacement volume is equal to the volume of air/fuel mixture sucked into the cylinder, because as the piston moves back, thus enlarging the combustion chamber volume, a negative pressure is created.

Due to the outside air pressure being higher, air is pushed into the chamber until both pressures are equal again. From this can be concluded that the displacement of the piston is equal to the volume of air sucked in. This air rushing into the combustion brings fuel with it, while the turbulent air also mixes the fuel equal throughout the air in the chamber. From this volume of air/fuel mixture, the maximum energy output from one stroke can be calculated.

To start, the cylinder bore diameter is equal to the diameter of the piston, 35.25 mm. The distance the piston moves during one stroke is 50 mm. So to calculate the displacement the surface area of the piston has to be multiplied by the travel of the piston. In cm^3, this would be: (pi*(3.

525/2)^2) * 5 which is 48.8 cm^3 or 48.8 cc to put it into standard units when talking about °displacement volumes, measured in cc (1cc = 1ml)The stoichiometric, or optimal, fuel to air ratio for gasoline is about 15 grams of air for 1 gram of fuel.The reaction would only create CO2 and H2O following this equation:25 O2 + 2 C8H18 ? 16 CO2 + 18 H2O Thus, with optimal combustion parameters of having 15 grams of air for 1 gram of fuel, the amount of fuel in 48.8 cm^3 of air can be calculated.Density of air at 15 °C at sea level is 1.

225 kg/m^3, so in 48.8 ml would 1.225*1000/(100^3)*48.8 = 6.0*10^-2 grams of air.To form an optimal combustion, (6.

0*10^-2)/15 grams of fuel would be needed. This is 4.0*10^-3 grams of gasoline.

The average energy content of gasoline is 46.7 MJ/kg, so 4.0*10^-3 grams of gasoline contains 1.86*10^-4 MJ of energy = 186 J with each ignition in our engine’s case.For each 2 moles of gasoline burned, 16 moles of CO2 as well as 18 moles of H2O are created, as can be seen in the reaction equation above. Due to the temperature generated by the exothermic combustion of gasoline, the H2O is formed in the gaseous state. CO2 is too. So for each 25 moles of O2 gas, 34 moles of gas are produced.

This is an increase of 25/34 = 1.36 times. The gasoline volume is neglected because the gasoline is in a liquid state and thus has a very small volume.Due to the temperatures occurring during a combustion, the volume of the gasses increases even further. With temperatures rising up to 600C° -700C°, there is a massive increase in volume.Following the gas laws, volume is directly proportional to the temperature. So just before combustion, the temperature can be as low as 80C°, while just after combustion temperatures can be as high as 700°C. In °K is this 353.

15°K to 973.15°K. This would be an increase of up to three times the temperature, so the volume of the gasses would also rise up to three times the original volume.Add to this that the engine compresses the 48.8 ml + the dead space above the piston of air into only the space that was the dead space above the piston. In our case, with a compression ratio of about 8:1, 7 mm of dead space is left above the piston when it is at its highest point, also called Top Dead Center or TDC for short. Pushing 48.8 ml of air into a volume of (pi(3.

525/2)^2)*0.7 = 6.8 ml, the pressure will also increase dramatically. This would increase by as much as the volume changes, so the total volume a p=0, is 48.8 + 6.8 ml = 55.

6 ml. This gets pushed into a volume of 6.8 ml, thus rising the pressure 55.6/6.8= 8 times.

All these rises in pressure together could form an increase of pressure of up to 1.36*3*8 = 32.64 times.So starting a p=0, = 1 atm = 1 bar, the eventual pressure could be as high as ~32 bars. The engine has to be designed to be able to hold all this pressure.

If the parts used are not strong enough, those parts will bend or break and ultimately not be able to perform their function to the fullest.